Hi yalexa10-ga,
At the beginning, when t = 0, the position of the wrench is 1000ft
above ground. Therefore, the position function would be:
s=1000-16t^2 (where 's' is feet above ground level)
At the moment of impact, s=0, which gives:
0=1000-16t^2
Rearranging this gives:
16t^2=1000
then:
t^2=1000/16
then:
t=(1000/16)^0.5
In other words, t equals the square root of 1000/16, or 7.90569 seconds.
To find the velocity at impact we consider the velocity to be the
limit (as 't' approaches the time of impact 'a') of:
v = ( s(a) - s(t) ) / (a-t)
Substituting the position function gives:
v = ( 1000-16a^2 - (1000-16t^2) ) / (a-t)
which is equivalent to:
v = (1000 - 16a^2 - 1000 + 16t^2) / (a-t)
which is equivalent to:
v = 16(t^2-a^2)/(a-t)
It is an algebraic identity, true for any 't' and 'a', that t^2-a^2 is
equal to (t+a)(t-a), which is equal to -(a+t)(a-t). Substituting this
gives:
v = 16(a+t)(a-t)/(a-t)
The two occurrences of (a-t) cancel out, leaving us with:
v = 16(a+t)
In the limit, as 't' approaches the time of impact 'a', this is equivalent to
v = 16(a+a)
Our time of impact is 7.90569 seconds, therefore the velocity at impact must be:
16(7.90569+7.90569)
which is 252.982 feet per second.
I trust this provides the information you require. If any of the steps
are not clear, please request clarification.
Regards,
eiffel-ga
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One-dimensional kinematics
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